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Sketch function from derivatives and limits | Capital One Interview Question

Q: Sketch function from derivatives and limits

This question evaluates understanding of calculus concepts—derivatives, concavity, limits, and continuity—and the ability to interpret derivative sign patterns to infer monotonicity and locate extrema.

Function analysis with derivative and concavity constraints

Context (clarified): Assume f: R → R is differentiable everywhere and twice differentiable on R \ {2}. The second derivative f'' exists and has the stated signs on open intervals; f is continuous at x = 2 where f'' is undefined.

Given:

Limits: lim_{x→−∞} f(x) = 0 and lim_{x→+∞} f(x) = 5.
First derivative:
- f'(x) > 0 on (−∞, −2) and (1, 3).
- f'(x) < 0 on (−2, 1) and (3, ∞).
- f'(x) = 0 only at x ∈ {−2, 1, 3}.
Second derivative:
- f''(x) < 0 on (−∞, −1) and (−1, 2).
- f''(x) > 0 on (2, ∞).
- f'' is undefined at x = 2, but f is continuous everywhere.

Tasks: (a) Sketch a qualitatively correct graph of f, marking all local extrema and any inflection points. (b) Is it possible for f to have two local maxima and one local minimum consistent with the above? Justify rigorously. (c) Determine where an inflection point must occur, if any, and explain the sign of the slope immediately before and after x = 2.

Function analysis with derivative and concavity constraints

Given:

Limits: lim_{x→−∞} f(x) = 0 and lim_{x→+∞} f(x) = 5.

First derivative:

f'(x) > 0 on (−∞, −2) and (1, 3).
f'(x) < 0 on (−2, 1) and (3, ∞).
f'(x) = 0 only at x ∈ {−2, 1, 3}.

Second derivative:

f''(x) < 0 on (−∞, −1) and (−1, 2).
f''(x) > 0 on (2, ∞).
f'' is undefined at x = 2, but f is continuous everywhere.

Sketch function from derivatives and limits

Quick Overview

Function analysis with derivative and concavity constraints

Solution

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Sketch function from derivatives and limits

Quick Overview

Function analysis with derivative and concavity constraints

Solution

Comments (0)